Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{-q^3 - 5q^2 - 4q}{9q^3 - 9q^2 - 18q}$
First factor out the greatest common factors in the numerator and in the denominator. $ a = \dfrac {-q(q^2 + 5q + 4)} {9q(q^2 - q - 2)} $ $ a = -\dfrac{q}{9q} \cdot \dfrac{q^2 + 5q + 4}{q^2 - q - 2} $ Simplify: $ a = - \dfrac{1}{9} \cdot \dfrac{q^2 + 5q + 4}{q^2 - q - 2}$ Since we are dividing by $q$ , we must remember that $q \neq 0$ Next factor the numerator and denominator. $ a = - \dfrac{1}{9} \cdot \dfrac{(q + 1)(q + 4)}{(q + 1)(q - 2)}$ Assuming $q \neq -1$ , we can cancel the $q + 1$ $ a = - \dfrac{1}{9} \cdot \dfrac{q + 4}{q - 2}$ Therefore: $ a = \dfrac{ -q - 4 }{ 9(q - 2)}$, $q \neq -1$, $q \neq 0$